3.6 Concentration Units

When solid solutes are dissolved in water, their molecules or ions readily move about in solution. They come in contact with one another readily in solution, so they combine easily. The quantity of a solute present in a given amount of solvent or solution is called the concentration of a solution.

Recall that you have learned the difference between a dilute and a concentrated solution. A dilute solution is that whose concentration is relatively low while a concentrated solution is a solution whose concentration is relatively high. You can find dilute and concentrated solutions of common acids and bases in your chemistry laboratory. Chemists use many concentration units. Here we will discuss molarity and strength.

i. Molarity (M)(mol/dm3)

If you read the label on the bottle of concentrated ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$$\mathrm{H}_{2}\mathrm{SO}_{4}$ you will notice $98\%$ by mass and also 18M. What does 18M stand for? This means there are 18 moles of ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$$\mathrm{H}_{2}\mathrm{SO}_{4}$ in each dm3 of solution. Similarly, conc. HCl is $37\%$ and 12.1 M HCl. This means there are 12.1 moles of HCl in each dm3 of solution. We can express the concentration in terms of moles of solute in the given volume of solution.

Molarity is the concentration unit in which the amount of solute is expressed in moles and the quantity of solution in dm3.

Molarity is defined as the number of moles of solute dissolved per dm3 of solution. Mathematically,

3.6.1 Problems Involving the Molarity of a Solution

Example 3.8

Urea is a white solid used as fertilizer and starting material for synthetic plastic. A solution contains 40g urea dissolved in $500{\mathrm{c}\mathrm{m}}^3$$500~\mathrm{cm}^3$ of solution. Calculate the molarity of this solution.

Problem Solving Strategy

To calculate the molarity, you need the number of moles of solute (which are not given) and the volume of solution in dm3. Convert the volume of solution $500{\mathrm{c}\mathrm{m}}^3$$500~\mathrm{cm}^3$ to dm3. Determine moles of solute from its mass using its molar mass.

Solution

Mass of urea $= 40g$

Molar mass of urea $=14+1\times 2+12+16+14+1\times 2$$= 14 + 1\times 2 + 12 + 16 + 14 + 1\times 2$ $= 60g / mol$

Moles of urea $\mathbf{\Sigma }=\frac{40}{60}=0.667\mathrm{m}\mathrm{o}\mathrm{l}$$\mathbf{\Sigma} = \frac{40}{60} = 0.667\mathrm{mol}$

Volume of solution $=500c{m}^3=\frac{500}{1000}=0.5d{m}^3$$= 500cm^3 = \frac{500}{1000} = 0.5dm^3$

1.334M

Example 3.9

Calculating molarity from moles of solute and volume of solution.

Potassium permanganate $(\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4)$$\mathrm{(KMnO_4)}$ is a dark blue- black compound. When it is dissolved in water, it forms a bright purple solution. It is used as a disinfectant in water tanks. It is also known as pinky. A solution contains 0.05 moles of $(\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4)$$\mathrm{(KMnO_4)}$ in $600{\mathrm{c}\mathrm{m}}^3$$600\mathrm{cm}^3$ of solution. Calculate the molarity of this solution.

Problem Solving Strategy

To calculate molarity, you need moles of solute which are given, and the volume of solution in dm. But volume is given in cm. So convert the cm to dm by dividing with 1000. Use a formula to calculate molarity.

Solution

$$\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}(\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4)=0.05$$$\mathrm{Moles~of~(KMnO_4)} = 0.05$ $$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=600\mathrm{c}{\mathrm{m}}^3$$$\mathrm{Volume~of~solution} = 600~\mathrm{cm^3}$ $$=\frac{600}{1000}$$$\mathrm{~ = \frac{600}{1000}}$ $$=0.6\mathrm{d}{\mathrm{m}}^3$$$\mathrm{~ = 0.6~dm^3}$

Now

$$\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}=\frac{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{{\mathrm{d}\mathrm{m}}^3\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}$$$\mathrm{Molarity} = \frac{\mathrm{moles~of~solute}}{\mathrm{dm}^3\mathrm{~of~solution}}$ $$\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}=\frac{0.05}{0.6}$$$\mathrm{Molarity} = \frac{0.05}{0.6}$ $$=0.083\mathrm{M}$$$\mathrm{~ = 0.083M}$

Concept Assessment Exercise 3.9

Potassium chlorate $(\mathrm{K}\mathrm{C}\mathrm{l}{\mathrm{O}}_3)$$\mathrm{(KClO_3)}$ is a white solid. It is used in making matches and dyes. Calculate the molarity of the solution that contains. (a) 1.5 moles of this compound dissolved in $250{\mathrm{c}\mathrm{m}}^3$$250\mathrm{cm}^3$ of solution (b) 75g of this compound dissolved to produce $1.25{\mathrm{d}\mathrm{m}}^3$$1.25\mathrm{dm}^3$ of solution. (c) What is the molarity of a $50{\mathrm{c}\mathrm{m}}^3$$50\mathrm{cm}^3$ sample of potassium chlorate solution that yields 0.25g residue after evaporation of the water.

1i. Strength of a Solution

The strength of a solution refers to the concentration of solute in a given volume of solution. It is expressed in terms of grams of solute per unit volume of solution. It is typically represented as:

g/dm (grams per cubic decimeter) or

g/cm (grams per cubic centimeter).

These units express how much solute is dissolved in a specified volume of solution. Mathematically

Mass of solute is measured in grams (g). Volume of solution is measured in cubic decimeters $(\mathrm{d}{\mathrm{m}}^3)$$\mathrm{(dm^3)}$ or cubic centimeters $(\mathrm{c}{\mathrm{m}}^3)$$\mathrm{(cm^3)}$

Unit Conversion:

$$1{\mathrm{d}\mathrm{m}}^3=1000{\mathrm{c}\mathrm{m}}^3$$$1\mathrm{dm}^3 = 1000\mathrm{cm}^3$

So, to convert between g/dm and g/cm:

$$1\mathrm{g}\mathrm{/}\mathrm{d}{\mathrm{m}}^3=0.001\mathrm{g}\mathrm{/}\mathrm{c}{\mathrm{m}}^3$$$1\mathrm{g / dm^3} = 0.001\mathrm{g / cm^3}$

Example 3.10 (g/dm)

A solution contains 20 g of salt dissolved in 2 dm of solution. Calculate the strength of the solution.

Solution

$$\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}(\mathrm{g})=20\mathrm{g}$$$\mathrm{Mass~of~solute~(g)} = 20\mathrm{g}$ $$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=2{\mathrm{d}\mathrm{m}}^3$$$\mathrm{Volume~of~solution} = 2\mathrm{dm}^3$ $$\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}(\mathrm{g})}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}(\mathrm{d}{\mathrm{m}}^3)}$$$\mathrm{Strength~of~solution} = \frac{\mathrm{Mass~of~solute~(g)}}{\mathrm{Volume~of~solution~(dm^3)}}$ $$\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=\frac{20\mathrm{g}}{2{\mathrm{d}\mathrm{m}}^3}$$$\mathrm{Strength~of~solution} = \frac{20\mathrm{g}}{2\mathrm{dm}^3}$ $$=10\mathrm{g}\mathrm{/}\mathrm{d}{\mathrm{m}}^3$$$= 10\mathrm{g / dm^3}$

Example 3.11 (g/cm)

A solution contains 25g of sugar dissolved in $500{\mathrm{c}\mathrm{m}}^3$$500\mathrm{cm}^3$ of solution. Calculate the strength of the solution.

Solution

$$\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}=25\mathrm{g}$$$\mathrm{Mass~of~solute} = 25\mathrm{g}$ $$\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=500{\mathrm{c}\mathrm{m}}^3$$$\mathrm{Volume~of~solution} = 500\mathrm{cm}^3$ $$\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}(\mathrm{g})}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}(\mathrm{c}{\mathrm{m}}^3)}$$$\mathrm{Strength~of~solution} = \frac{\mathrm{Mass~of~solute~(g)}}{\mathrm{Volume~of~solution~(cm^3)}}$ $$\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=\frac{25\mathrm{g}}{500{\mathrm{c}\mathrm{m}}^3}$$$\mathrm{Strength~of~solution} = \frac{25\mathrm{g}}{500\mathrm{cm}^3}$ $$=0.05\mathrm{g}\mathrm{/}\mathrm{c}{\mathrm{m}}^3$$$= 0.05\mathrm{g / cm^3}$

3.6.2 Problems Involving Interconversion of Molarity and Strength Example 3.13

Converting the molarity of a solution into its concentration in $\mathrm{g}\mathrm{/}{\mathrm{d}\mathrm{m}}^3$$\mathrm{g} / \mathrm{dm}^{3}$ and $\mathrm{g}\mathrm{/}{\mathrm{c}\mathrm{m}}^3$$\mathrm{g} / \mathrm{cm}^{3}$ A flask contains $0.25\mathrm{M}$$0.25\mathrm{M}$ NaOH solution. What mass of NaOH is present per ${\mathrm{d}\mathrm{m}}^3$$\mathrm{dm}^{3}$ of solution? Problem Solving Strategy Molarity means the number of moles per ${\mathrm{d}\mathrm{m}}^3$$\mathrm{dm}^{3}$ of solution. So, $0.25\mathrm{M}$$0.25\mathrm{M}$ NaOH means 0.25 moles of NaOH dissolved per ${\mathrm{d}\mathrm{m}}^3$$\mathrm{dm}^{3}$ of solution. You need to convert moles of solute to mass using molar mass. Solution Molarity $\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}=0.25\mathrm{M}$$\mathrm{Moles~of~NaOH} = 0.25\mathrm{M}$ Moles of NaOH $= 0.25$ Volume of solution $=1{\mathrm{d}\mathrm{m}}^3$$= 1\mathrm{dm}^{3}$ Molar mass of NaOH $= 23 + 16 + 1$ $=40\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$$= 40\mathrm{g} / \mathrm{mole}$ Moles of solute $=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{e}(\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e})}$$= \frac{\mathrm{mass~of~solute}}{\mathrm{molar~mass~of~solute(g / mole)}}$ $0.25\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}{40\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}}$$0.25\mathrm{moles} = \frac{\mathrm{mass~of~NaOH}}{40\mathrm{g} / \mathrm{mole}}$ Mass of NaOH $=0.25\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\times 40\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$$= 0.25\mathrm{moles}\times 40\mathrm{g / mole}$ $=10\mathrm{g}$$= 10\mathrm{g}$ Thus, the solution contains $10\mathrm{g}$$10\mathrm{g}$ NaOH per ${\mathrm{d}\mathrm{m}}^3$$\mathrm{dm}^{3}$ Now As $1{\mathrm{d}\mathrm{m}}^3=1000{\mathrm{c}\mathrm{m}}^3$$1\mathrm{dm}^{3} = 1000\mathrm{cm}^{3}$ The solution contains $10\mathrm{g}\mathrm{/}{\mathrm{d}\mathrm{m}}^3$$10\mathrm{g} / \mathrm{dm}^{3}$ of solution or the solution contains $10\mathrm{g}$$10\mathrm{g}$ of NaOH per $1000{\mathrm{c}\mathrm{m}}^3$$1000\mathrm{cm}^{3}$ So, $1000{\mathrm{c}\mathrm{m}}^3$$1000\mathrm{cm}^{3}$ of solution contains $=10\mathrm{g}$$= 10\mathrm{g}$ NaOH $1{\mathrm{c}\mathrm{m}}^3$$1\mathrm{cm}^{3}$ of solution contains $=10\mathrm{g}\mathrm{/}1000$$= 10\mathrm{g} / 1000$ $=0.01\mathrm{g}$$= 0.01\mathrm{g}$ NaOH Therefore concentration of solution is $=0.01\mathrm{g}\mathrm{/}{\mathrm{c}\mathrm{m}}^3$$= 0.01\mathrm{g} / \mathrm{cm}^{3}$

3.14

Converting concentration in $\mathrm{g}\mathrm{/}{\mathrm{d}\mathrm{m}}^3$$\mathrm{g} / \mathrm{dm}^3$ into molarity.

Potassium hydroxide (KOH) is used in the manufacturing of shaving creams, paints and varnish. An analyst makes up a solution by dissolving $5.8\mathrm{g}$$5.8\mathrm{g}$ of KOH in one dm $^3$ of solution. Calculate the molarity of this solution.

Problem solving strategy

To calculate the molarity, you need moles of solute per dm $^3$ of solution. Moles of solute are not given. But the mass of solute per dm $^3$ of the solution is given, Convert the mass of the solute into moles by using its molar mass.

Solution

Mass of KOH dissolved in one dm $^3$ of solution $=5.6\mathrm{g}$$= 5.6\mathrm{g}$

Molar mass of KOH $= 39 + 16 + 1$

$$=56\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}$$$= 56\mathrm{g / mol}$

Moles of KOH $\mathrm{\Omega }=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{K}\mathrm{O}\mathrm{H}(\mathrm{g})}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{K}\mathrm{O}\mathrm{H}(\mathrm{g})}$$\mathrm{\Omega} = \frac{\mathrm{mass~of~KOH(g)}}{\mathrm{molar~mass~of~KOH(g)}}$

$$=\frac{5.6}{56}$$$= \frac{5.6}{56}$

Thus, the solution contains 0.1 moles of KOH in one dm $^3$ of the solution, so the molarity of the solution is 0.1M.

Teacher's Point

A teacher may give different concentration interconversion numerical to students as homework.

Concept Assessment Exercise 3.10

1. Sodium hydroxide solutions are used to neutralize acids and in the preparation of soaps and rayon. If you dissolve 25g of NaOH to make 1 dm $^3$ of solution, what is the molarity of this solution?

2. A solution of NaOH has a concentration of 1.2M. Calculate the mass of NaOH in $\mathrm{g}\mathrm{/}{\mathrm{d}\mathrm{m}}^3$$\mathrm{g} / \mathrm{dm}^3$ and $\mathrm{g}\mathrm{/}{\mathrm{c}\mathrm{m}}^3$$\mathrm{g} / \mathrm{cm}^3$ in this solution.

3. A solution is prepared by dissolving 10g of haemoglobin in enough water to make up 1dm $^3$ in volume. Calculate molarity of this solution. Molar mass of haemoglobin is $6.51\times {10}^4\mathrm{g}\mathrm{/}\mathrm{m}\mathrm{o}\mathrm{l}$$6.51\times 10^{4}\mathrm{g / mol}$

3.6.3 Dilution of Solutions